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A lab of Reduction/Oxidation Reactions

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  • Pages: 5
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  • Category: Chemistry

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Research Question: Is it possible to determine if a redox reaction took place by using only the equation?

Hypothesis: Yes, I believe it is possible to determine if this reaction took place by using the oxidation numbers in the equation.


1. I cleaned the reaction surface to limit contamination

2. I held the chemicals at approximately a 90 degree angle for consistency


Redox reactions occur when both oxidation and reduction take place. All atoms can be assigned an oxidation number, which is a positive or negative number that help in determining the reducing and/or oxidizing agents in an equation. An increase in the oxidation number of an atom from one side of an equation to another side indicates oxidation. A decrease in the oxidation number of an atom indicates reduction. For example:

2AgNO3 + Cu Cu(NO3)2 + 2Ag

In this equation, on the left side of the equation the Ag has an oxidation number of silver is +1 and on the other side the Ag oxidation number is 0. Since the silver ions decrease in oxidation number, reduction is occurring. Similarly on the left hand of the equation the Cu has an oxidation number of 0 and on the side it has an oxidation number of +2. Since the copper ions increase in oxidation number, oxidation is taking place.


1. Reaction surface

2. Chemicals

a. Cl2, KMnO4, H2O2, KIO3, K2Cr2O7, NaNO2, CuSO4, FeCl3, KI, starch, HCl, NaHSO3, Na2S2O3

3. Pipette for stirring chemicals in reaction

4. Pencil and paper

5. Safety goggles*


1. Make a chart to record collected data from the experiment

2. Gather all necessary chemicals and other materials

3. Mix one drop of KI with one drop of NaOCl and observe if a reaction takes place

4. If no reaction is visible add one drop of starch to the mixture and once again observe if a reaction occurs. If a reaction occurs still add one drop of starch to the mixture to confirm results.

5. If a reaction did not occur between the chemicals or even with the starch add one drop of HCl and observe the reaction that occurs.

6. Add one drop of NaHSO3 to the mixture and observe the reaction.

7. Repeat steps 3-6 with the other seven chemical combinations.

8. However, in reactions e-h, instead of NaHSO3, add one drop of Na2S2O3.

9. Clean up the lab area and wash your hands.*


KI Starch HCl NaHSO3 Na2S2O3

Cl2 Yellow Purple-black Orange w/ brown precipitate

KMnO4 Yellow w/ brown precipitate Purple-black Yellow w/ brown precipitate

H2O2 Light yellow Purple-black Light yellow

KIO3 Clear (no rxn) Clear (no rxn) Purple-black Purple-black

K2Cr2O7 Orange- yellow (no rxn) Orange-yellow (no rxn) Purple-black Orange yellow

NaNO2 Clear (no rxn) Clear (no rxn) Purple-black Clear

CuSO4 Brown Purple-black Light green

FeCO3 Yellow (no rxn) Purple-black Purple-black

In all of these reactions oxidation and reduction took place because iodine visibly came out of the solutions in a purple form. In some of the reactions extra H+ ions were needed to balance out the oxygen atoms in the solutions. These extra H+ ions were probably given by the addition of HCl. Also, NaHSO3 and Na2S2O3 were used to reverse the reactions, which is what happened in the majority of our solutions. However, in some of the solutions the addition of these chemicals did not reverse the reaction. This could be due to a number of reasons but the most probable I believe is contamination.


a. Cl2 + I­ Cl- + I2

Cl2 + 2e- 2Cl- reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

Cl2 + 2I- 2Cl- + I2 final equation

b. MnO4- + I- Mn2+ + I2

MnO4- + 5e- Mn2+ reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

2MnO4- +10I- 2Mn2+ + 5I2 final equation

c. H2O2 +I- H2O +I2

H2O2 + e- H2O reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

2H2O2 +2I- 2H2O + I2 final equation

d. IO3- + I- I- +I2

IO3- + 6e- I- reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

IO3- + 6I- 3I2 + I- final equation

e. Cr2O72- + I- Cr3+ + I2

Cr2O72- + 6e- 2Cr3+ reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

Cr2O72- + 6I- 3I2 + 2Cr3+ final equation

f. NO2- + I- NO + I2

NO2- + 3e- NO reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

2NO2- + 6I- 2NO- + 3I2 final equation

g. Cu2+ + I- CuI + I2

Cu2+ + 2e- CuI reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

Cu2+ +2I- I2 + CuI final equation

h. Fe3+ + I- Fe2+ + I2

Fe3+ +e- Fe2+ reduced oxidizing agent

2I- I2 + 2e- oxidized reducing agent

2Fe3+ + 2I- 2Fe2+ + I2 final equation

In the first problem, a, I wrote out the equation given in the experiment. I then assigned oxidation numbers to all of the atoms. Cl2 is 0 because it is a neutral compound and I- is negative 1 because it is equal to the magnitude and sign of its ionic charge. On the other side Cl- has an oxidation number of -1 because of its magnitude and sign of its ionic charge and I2 has a number of 0 because it is neutral. Since Cl2 gains electrons, it is reduced and since I- loses electrons it is oxidized. All of the other equations were solved in a similar fashion.


In this experiment my hypothesis was correct; it is possible to determine if a reaction is a redox reaction based on oxidation numbers. In my analysis I wrote out the equations for all of the reactions and using the oxidation numbers I found that both oxidation and reduction processes were occurring in all of them. Previously, I mixed solutions together to see if a reaction would take place. In all of these reactions, oxidation and reduction were taking place because of the appearance of iodine coming out of the solution. All of the data I collected from the reactions agreed with the equations that I wrote out at the end, that redox reactions were occurring with iodine as the reducing agent and the other chemicals as the oxidizing agents.

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